Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $t = \dfrac{-3y^2 - 30y}{y^3 + 8y^2 + 16y} \div \dfrac{y + 6}{y^2 + 10y + 24} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $t = \dfrac{-3y^2 - 30y}{y^3 + 8y^2 + 16y} \times \dfrac{y^2 + 10y + 24}{y + 6} $ First factor out any common factors. $t = \dfrac{-3y(y + 10)}{y(y^2 + 8y + 16)} \times \dfrac{y^2 + 10y + 24}{y + 6} $ Then factor the quadratic expressions. $t = \dfrac {-3y(y + 10)} {y(y + 4)(y + 4)} \times \dfrac {(y + 4)(y + 6)} {y + 6} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {-3y(y + 10) \times (y + 4)(y + 6) } { y(y + 4)(y + 4) \times (y + 6)} $ $t = \dfrac {-3y(y + 4)(y + 6)(y + 10)} {y(y + 4)(y + 4)(y + 6)} $ Notice that $(y + 4)$ and $(y + 6)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {-3y\cancel{(y + 4)}(y + 6)(y + 10)} {y\cancel{(y + 4)}(y + 4)(y + 6)} $ We are dividing by $y + 4$ , so $y + 4 \neq 0$ Therefore, $y \neq -4$ $t = \dfrac {-3y\cancel{(y + 4)}\cancel{(y + 6)}(y + 10)} {y\cancel{(y + 4)}(y + 4)\cancel{(y + 6)}} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $t = \dfrac {-3y(y + 10)} {y(y + 4)} $ $ t = \dfrac{-3(y + 10)}{y + 4}; y \neq -4; y \neq -6 $